12.33 Cotangent extensions with zero ruling componentC9CE007F34A7

Proof. As in the proof of lemma 12.31.1, holomorphy in the infinite ruling chart bounds the degree in $t$ of the coefficient of $y^ j$ by $2j$; the variable $w$ does not change between the charts. Distributing the exponent $a$ among $j$ integers in $\{ 0,1,2\} $, every monomial $t^ ay^ j$ with $0\leq a\leq 2j$ is a product of $y$, $ty$ and $t^2y$. These generators satisfy $y(t^2y)=(ty)^2$.

In degree $j$, the space of functions of $y$ and $t$ has dimension $2j+1$. The polynomial algebra in three variables modulo the nonzero quadratic relation has the same dimension

where the second term is zero for $j<2$, so the natural surjective graded map is an isomorphism. Adjoining the two trivial fiber coordinates $u,w$ gives the stated algebra, with entire coefficients in $z$. In the other chart,

so all generators are holomorphic there. Substitution gives their lattice transformations. $\square$

Proof. Give $c_-,c_+$ weights $-1,1$, and $h,u,w$ weight zero. Using $c_-c_+=h^2$, the blocks of negative weight have the form $c_-^ kF(z,h,u,w)$, those of positive weight the form $c_+^ kF(z,h,u,w)$, and the block of weight zero consists of the polynomials $F$. These expressions are unique: the substitution $c_+=h^2/c_-$ embeds the algebra into $\mathcal O(\mathbb C)[h,u,w,c_-,c_-^{-1}]$, and distinct weights correspond to distinct powers of $c_-$ with nonzero polynomial coefficients. In tensor degree $r$, the remaining polynomial degree is $r-k$.

The coefficient bundle $U_3$ now has the inverse translation $u\mapsto u-m(q_0h+Kw)$. Thus the block of weight $-k$ is $H^0(E,A^ k\otimes \operatorname {Sym}^{r-k}U_3)$, the block of weight $k$ has $A^{-k}$ in place of $A^ k$, and the block of weight zero is $H^0(E,\operatorname {Sym}^ rU_3)$. Put $H=h+(K/q_0)w$. Since $q_0\ne 0$, this identifies $U_3$ with the bundle $U$ defined above, with $H$ in place of its Euler variable, plus the trivial line bundle generated by $w$.

By lemma 12.31.2, all positive weights vanish, and the algebra of weight zero is $\mathbb C[H,w]$. Expanding the negative blocks in powers of $w$ identifies them with the blocks for the tangent bundle in lemma 12.31.1, and the substitutions $h\mapsto H$ and $y^ k\mapsto c_-^ k$ give bijections in every degree and surviving weight. Products multiply the coefficient polynomials and add the powers of $c_-$, so these bijections preserve multiplication. This proves the isomorphism. Summing the dimensions for the tangent bundle and taking the quotient by $w$ gives the remaining assertions, exactly as in the proof of corollary 12.32.3. $\square$

Proof. Set $w=1$. The subset $\{ y=0\} $ of the finite ruling chart agrees with $\{ y_\infty =0\} $ in the other chart, and the lattice acts on it by $u\mapsto u+Km$. It therefore descends to a submanifold of $T_ W$ that is a $\mathbb P^1$-bundle over

this quotient is a complex manifold because translation in $z$ gives a free and properly discontinuous action. Every fiber of this bundle is a compact connected curve in $T_ W$, on which holomorphic functions are constant. Thus $T_ W$ is not holomorphically separable, and hence not Stein.

Before the lattice quotient, the affine charts have no affine term in the ruling direction, and they glue to $\mathbb C^2_{z,u}\times \operatorname {Tot}(\mathcal O_{\mathbb P^1}(-2))$, since $\mathcal O_{\mathbb P^1}(-2)$ is the cotangent bundle of $\mathbb P^1$. This space is simply connected: contracting the fibers of the line bundle retracts the last factor onto $\mathbb P^1$, and $\mathbb C^2$ is contractible. The lattice acts freely and properly discontinuously, so this space is the universal cover. Its zero section contains a compact projective line, which again obstructs holomorphic separability and hence the Stein property. $\square$